What is the derivative of h(x)=\sin2x\cos2x

nemired9 2021-12-26 Answered
What is the derivative of \(\displaystyle{h}{\left({x}\right)}={\sin{{2}}}{x}{\cos{{2}}}{x}\)

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Expert Answer

Orlando Paz
Answered 2021-12-27 Author has 716 answers
Step 1
Product Rule of differentiation:
If f(x) and g(x) are any two differentiable functions, then
\(\displaystyle{f{{\left({g}\right)}}}'{\left({x}\right)}={f{{\left({x}\right)}}}\times{g}'{\left({x}\right)}+{f}'{\left({x}\right)}\times{g{{\left({x}\right)}}}\)
Chain rule of differentiation:
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{f{{\left({g{{\left({x}\right)}}}\right)}}}={f}'{\left({g{{\left({x}\right)}}}\right)}\times{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{g{{\left({x}\right)}}}\)
\(\displaystyle={f}'{\left({g{{\left({x}\right)}}}\right)}\times{g}'{\left({x}\right)}\times{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}\right)}\)
\(\displaystyle={f}'{\left({g{{\left({x}\right)}}}\right)}\times{g}'{\left({x}\right)}\)
The given function is \(\displaystyle{h}{\left({x}\right)}={\sin{{2}}}{x}{\cos{{2}}}{x}\)
Step 2
Differentiate the given function using the product rule as follows.
\(\displaystyle{h}'{\left({x}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{2}}}{x}{\cos{{2}}}{x}\right)}\)
\(\displaystyle={\sin{{2}}}{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\cos{{2}}}{x}\right)}+{\cos{{2}}}{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{2}}}{x}\right)}\)
(U sin g product rule)
\(\displaystyle={\sin{{2}}}{x}{\left(-{\sin{{2}}}{x}\right)}{\left({2}\right)}+{\cos{{2}}}{x}{\left({\cos{{2}}}{x}\right)}{\left({2}\right)}\)
(U sin g chain rule)
\(\displaystyle=-{2}{{\sin}^{{{2}}}{2}}{x}+{2}{{\cos}^{{{2}}}{2}}{x}\)
\(\displaystyle=-{2}{\left({1}-{{\cos}^{{{2}}}{2}}{x}\right)}+{2}{{\cos}^{{{2}}}{2}}{x}\)
\(\displaystyle=-{2}+{4}{{\cos}^{{{2}}}{2}}{x}\)
Therefore, the derivative of the given function is
\(\displaystyle{h}'{\left({x}\right)}=-{2}+{4}{{\cos}^{{{2}}}{2}}{x}\)
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peterpan7117i
Answered 2021-12-28 Author has 2576 answers
Step 1
Given function:
\(\displaystyle{\sin{{\left({2}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}\)
\(\displaystyle{\frac{{{1}}}{{{2}}}}{\left({2}{\sin{{\left({2}{x}\right)}}}{\cos{{\left({2}{x}\right)}}}\right)}\)
\(\displaystyle{\frac{{{1}}}{{{2}}}}{\sin{{\left({4}{x}\right)}}}\)
Differentiating given function w.r.t x as follows
\(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\frac{{{1}}}{{{2}}}}{\sin{{\left({4}{x}\right)}}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\sin{{\left({4}{x}\right)}}}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\cos{{\left({4}{x}\right)}}}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({4}{x}\right)}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{\cos{{\left({4}{x}\right)}}}{\left({4}\right)}\)
\(\displaystyle={2}{\cos{{\left({4}{x}\right)}}}\)
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karton
Answered 2022-01-04 Author has 8659 answers

Expalnation:
differentiate using the \(\frac{\text{product}}{\text{chain rule}}\)
\(\begin{array}{} \text{Given} \\y=f(x)g(x) \\\frac{dy}{dx}=f(x)g'(x)+g(x)f'(x)\Rightarrow \text{product rule} \\f(x)=\sin2x\Rightarrow f'(x)=2\cos2x \\g(x)=\cos2x\Rightarrow g'(x)=-2\sin2x \\\frac{d}{dx}(\sin2x\cos2x) \\=-2\sin^{2}2x+2\cos^{2}2x \\=2(\cos^{2}2x-\sin^{2}2x)=2\cos4x \end{array}\)

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